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Curvature of a circle approaches zero as radius goes to infinity 
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#1
Apr1514, 06:33 AM

P: 271

Hello,
this isn't a homework problem, so i'm hoping it's okay to post here. I would like to know the correct way to mathematically express the idea in my title. It is intuitively obvious that as the radius of a circle increases, it's curvature decreases. I looked it up and found that the curvature of a circle is equal to the reciprocal of it's radius. Certain assumptions are often made when looking at lenses, i.e the wave fronts reaching the lens are parallel, or have 0 curvature  In other words, the object distance is infinitely far away. But, 1/∞ ≠ 0 So how do I express it properly? In words, I think it goes something like this  As the radius tends towards infinity, the curvature of the circle tends towards zero. 


#2
Apr1514, 06:53 AM

P: 2,990

Wouldn't you just use the lim 1/r expressions with r> infinity to express it?



#3
Apr1514, 07:01 AM

P: 271

[itex]lim_{r \rightarrow ∞} \frac{1}{r} = 0[/itex] Like that? 


#4
Apr1514, 07:20 AM

P: 2,990

Curvature of a circle approaches zero as radius goes to infinity
Yes thats the way I'd express it.



#5
Apr1614, 08:37 AM

P: 18

If you imagine a circle with infinite radius, then its circumference is also infinite.
Then what would be the value of pi be? Infinite divided by infinite. Can you say what it is? I think the real projective line may be a picture of this kind of "circle": http://en.wikipedia.org/wiki/Real_projective_line 


#6
Apr1614, 09:02 AM

Mentor
P: 21,286

They usually come up when we are evaluating limits of functions. 


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